Chapter 4, Problem 4.133AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to the reaction of ethanol and acetic acid used in making of apple cider vinegar are to be answered.

Concept introduction: The main component of apple cider vinegar is acetic acid. This vinegar helps to maintain heart health and blood pressure as well.

A compound is said to be oxidized when it loses electrons which increases its oxidation number. On the other hand reduction is the process of gaining electrons which decreases the oxidation number of the compound.

To determine: The balanced chemical equation describing the fermentation of natural sugars to ethanol and carbon dioxide.

Answer to Problem 4.133AP

Solution

The balanced chemical equation describing the fermentation of natural sugars to ethanol and carbon dioxide is,

$3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

Explanation of Solution

Explanation

The fermentation of natural sugars gives equal number of moles of ethanol and carbon dioxide.

The given empirical formula of natural sugar is ${\text{CH}}_{\text{2}}\text{O}$. The equation of fermentation of this sugar is,

${\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

The above equation is unbalanced. To balance the number of carbon, hydrogen and oxygen place coefficient of $3$ in front of ${\text{CH}}_{\text{2}}\text{O}$ to the reactant side of the equation as,

$3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

Therefore, the balanced chemical equation describing the fermentation of natural sugars to ethanol and carbon dioxide is,

$3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

(b)

Interpretation Introduction

To determine: The balanced chemical equation describing the acid fermentation of ethanol to acetic acid.

Answer to Problem 4.133AP

Solution

The balanced chemical equation describing the acid fermentation of ethanol to acetic acid is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

Explanation

The acid fermentation of ethanol $\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\right)$ to acetic acid $\left({\text{CH}}_{\text{3}}\text{COOH}\right)$ will take place in the presence of oxygen. This reaction is termed as oxidation reaction due to the reaction of ethanol with oxygen.

The chemical equation for the above fermentation is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The above equation consists of equal number of atoms on both the reactant side and the product side. Thus, there is no need to balance the above equation.

Therefore, the balanced chemical equation describing the acid fermentation of ethanol to acetic acid is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

(c)

Interpretation Introduction

To determine: The oxidation states of carbon in the reactants and the products of the two fermentation reactions.

Answer to Problem 4.133AP

Solution

In the first fermentation reaction, the oxidation number of carbon in ${\text{CH}}_{\text{2}}\text{O}$ is $\underset{\_}{0}$. the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is $\underset{\_}{+4}$.

In the second fermentation reaction, the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CH}}_3\text{COOH}$ is $\underset{\_}{0}$.

Explanation of Solution

Explanation

The first fermentation reaction is,

$3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

On the reactant side the given compound is ${\text{CH}}_{\text{2}}\text{O}$. The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation number of $\text{C}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on CH}}_{\text{2}}\text{O}=\left[\begin{array}{c}\left(\text{Number of C atoms}\times \text{oxidation number of C}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{CH}}_{\text{2}}\text{O}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on CH}}_{\text{2}}\text{O}=\left[\left(1\times x\right)+\left(2\times 1\right)+\left(1\times \left(-2\right)\right)\right]\\ 0=x+2+\left(-2\right)\\ 0=x+0\\ x=\underset{\_}{0}\end{array}$

Hence, the oxidation number of carbon in ${\text{CH}}_{\text{2}}\text{O}$ is $\underset{\_}{0}$.

On the product side the given element is ${\text{C}}_2{\text{H}}_5\text{OH}$. The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration. The oxidation number of $\text{C}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on C}}_2{\text{H}}_5\text{OH}=\left[\begin{array}{c}\left(\text{Number of C atoms}\times \text{oxidation number of C}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{C}}_2{\text{H}}_5\text{OH}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on C}}_2{\text{H}}_5\text{OH}=\left[\left(2\times x\right)+\left(6\times 1\right)+\left(1\times \left(-2\right)\right)\right]\\ 0=2x+6+\left(-2\right)\\ -4=2x\\ x=\underset{\_}{-2}\end{array}$

Hence, the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$.

The given element is ${\text{CO}}_{\text{2}}$. The oxidation state of oxygen in a compound is usually $-2$ and the oxidation state of carbon is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on CO}}_2=\left[\begin{array}{c}\left(\text{Number of C atoms}\times \text{oxidation number of C}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{CO}}_{\text{2}}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on CO}}_2=\left[\left(1\times x\right)+\left(2\times \left(-2\right)\right)\right]\\ 0=x-4\\ x=\underset{\_}{+4}\end{array}$

Hence, the oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is $\underset{\_}{+4}$.

Therefore, In the first reaction of fermentation, the oxidation number of carbon in ${\text{CH}}_{\text{2}}\text{O}$ is $\underset{\_}{0}$. the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is $\underset{\_}{+4}$.

The second fermentation reaction is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

On the reactant side the given element is ${\text{C}}_2{\text{H}}_5\text{OH}$. The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation number of $\text{C}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on C}}_2{\text{H}}_5\text{OH}=\left[\begin{array}{c}\left(\text{Number of C atoms}\times \text{oxidation number of C}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{C}}_2{\text{H}}_5\text{OH}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on C}}_2{\text{H}}_5\text{OH}=\left[\left(2\times x\right)+\left(6\times 1\right)+\left(1\times \left(-2\right)\right)\right]\\ 0=2x+6+\left(-2\right)\\ -4=2x\\ x=\underset{\_}{-2}\end{array}$

Hence, the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$.

On the product side the given element is ${\text{CH}}_3\text{COOH}$. The oxidation state of hydrogen in a compound is usually $+1$ as it contains only one valence electron and of oxygen is usually $-2$ as it requires two more electrons in order to achieve stable electronic configuration.

The oxidation number of $\text{C}$ is assumed to be $x$ and is calculated by using the formula,

${\text{Charge on CH}}_3\text{COOH}=\left[\begin{array}{c}\left(\text{Number of C atoms}\times \text{oxidation number of C}\right)+\\ \left(\text{Number of H atoms}\times \text{oxidation number of H}\right)+\\ \left(\text{Number of O atoms}\times \text{oxidation number of O}\right)\end{array}\right]$

Since, the sum of all oxidation numbers in a neutral compound is zero that is charge on ${\text{CH}}_3\text{COOH}$ is zero.

Substitute the number of atoms and their oxidation number in the above formula to calculate the oxidation number of carbon.

$\begin{array}{c}{\text{Charge on CH}}_3\text{COOH}=\left[\left(2\times x\right)+\left(4\times 1\right)+\left(2\times \left(-2\right)\right)\right]\\ 0=2x+4+\left(-4\right)\\ 0=2x\\ x=\underset{\_}{0}\end{array}$

Hence, the oxidation number of carbon in ${\text{CH}}_3\text{COOH}$ is $\underset{\_}{0}$.

Therefore, In the second reaction of fermentation, the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CH}}_3\text{COOH}$ is $\underset{\_}{0}$.

(d)

Interpretation Introduction

To determine: The maximum quantity of acetic acid that can be produce by the two fermentation reactions.

Answer to Problem 4.133AP

Solution

The maximum quantity of acetic acid that can be produce by the two fermentation reactions is $\underset{\_}{22.22g}$.

Explanation of Solution

Explanation

The first fermentation reaction is,

$3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$

The second fermentation reaction is,

${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Combine both the above reactions as,

${\text{3CH}}_2{\text{O+C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The overall equation of the above combined reactions is obtained by eliminating all the spectator elements as,

${\text{3CH}}_2\text{O}+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

This reaction is balanced which represents that $3$ moles of ${\text{CH}}_{\text{2}}\text{O}$ is producing $1$ mole of ${\text{CH}}_{\text{3}}\text{COOH}$.

The quantity of acetic acid that is produced from the above overall fermentation reaction is calculated as follows.

Firstly, the molar mass of ${\text{3CH}}_{\text{2}}\text{O}$ and ${\text{CH}}_{\text{3}}\text{COOH}$ is calculated as,

$\begin{array}{c}\text{Molar}\text{mass}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=\left(2\text{C+4H+2O}\right)\\ =\left(\begin{array}{c}2\times \text{12}\text{.01}\text{g}\text{C}+4\times 1.\text{01}\text{g}\text{H}\\ +2\times \text{16}\text{.00}\text{g}\text{O}\end{array}\right)\\ =\text{60}\text{.06}\text{g/mol}\end{array}$

$\begin{array}{c}\text{Molar}\text{mass}\text{of}3{\text{CH}}_2\text{O}=\left(\text{3C+6H+3O}\right)\\ =\left(\begin{array}{c}3\times \text{12}\text{.01}\text{g}\text{C}+6\times 1.\text{01}\text{g}\text{H}\\ +3\times \text{16}\text{.00}\text{g}\text{O}\end{array}\right)\\ =90.\text{05}\text{g/mol}\end{array}$

The given mass of natural sugar is $1.1\times {10}^2\text{g}$.

The moles of sugar ${\text{CH}}_{\text{2}}\text{O}$ is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

Substitute the values of mass and molar mass of sugar ${\text{CH}}_{\text{2}}\text{O}$ in the above formula as,

$\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_{\text{2}}\text{O}=\frac{1.1\times {10}^2\text{g}}{90.05\text{g/mol}}\\ =1.11\text{mol}\text{of}{\text{CH}}_{\text{2}}\text{O}\end{array}$

As $3$ moles of sugar ${\text{CH}}_{\text{2}}\text{O}$ is producing $1$ mole of ${\text{CH}}_{\text{3}}\text{COOH}$. Thus, total moles of ${\text{CH}}_{\text{3}}\text{COOH}$ is calculated by diving the moles of sugar by $3$ as,

$\begin{array}{c}\text{Moles}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=\frac{\text{Moles}\text{of}\text{sugar}{\text{CH}}_{\text{2}}\text{O}}{3}\\ =\frac{1.1\times {10}^2\text{g}}{3}\\ =0.37\text{mol}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\end{array}$

The quantity of acetic acid that is produced from the above overall fermentation reaction is calculated by the formula,

$\text{The}\text{yielded}\text{quantity}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\left(g\right)=\text{Molar}\text{mass}\times \text{Moles}$

Substitute the values of molar mass and moles of ${\text{CH}}_{\text{3}}\text{COOH}$ in the above formula,

$\begin{array}{c}\text{The}\text{yielded}\text{quantity}\text{of}{\text{CH}}_{\text{3}}\text{COOH}\left(g\right)=0.37\text{mol}{\text{CH}}_{\text{3}}\text{COOH}\times \frac{60.05\text{g}{\text{CH}}_{\text{3}}\text{COOH}}{1\text{mol}{\text{CH}}_{\text{3}}\text{COOH}}\\ =\underset{\_}{22.22gC{H}_{3}COOH}\end{array}$

Therefore, The maximum quantity of acetic acid that can be produce by the two fermentation reactions is $\underset{\_}{22.22g}$.

Conclusion

1. a) The balanced chemical equation describing the fermentation of natural sugars to ethanol and carbon dioxide is,

   $3{\text{CH}}_{\text{2}}\text{O}\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{CO}}_{\text{2}}\left(g\right)$.

2. b) The balanced chemical equation describing the acid fermentation of ethanol to acetic acid is,

   ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}\left(l\right)+{\text{O}}_{\text{2}}\left(g\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(l\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

3. c) In the first fermentation reaction, the oxidation number of carbon in ${\text{CH}}_{\text{2}}\text{O}$ is $\underset{\_}{0}$. the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CO}}_{\text{2}}$ is $\underset{\_}{+4}$.

In the second fermentation reaction, the oxidation number of carbon in ${\text{C}}_2{\text{H}}_5\text{OH}$ is $\underset{\_}{-2}$ and the oxidation number of carbon in ${\text{CH}}_3\text{COOH}$ is $\underset{\_}{0}$.

1. d) The maximum quantity of acetic acid that can be produce by the two fermentation reactions is $\underset{\_}{22.22g}$